# Math Help - Strange question

1. ## Strange question

An escalator takes 20 seconds to reach the next floor - if a child gets on the escalator he reaches the next floor in 4 seconds. How long would it take for the child to go up if the escalator isn't working?

Is it possible to work this out or is more information needed?

2. Originally Posted by sad_student
An escalator takes 20 seconds to reach the next floor - if a child gets on the escalator he reaches the next floor in 4 seconds. How long would it take for the child to go up if the escalator isn't working?

Is it possible to work this out or is more information needed?
Let the childs speed be b and escalators speed be a and the stair distance be d.
When the escalators moving up, the net speed of the boy is a+b.
Then $d = 20 a = 4(a+b) \Rightarrow 4a = b \Rightarrow d = 20a = 5b$

So the child takes 5 seconds to reach the next floor.

3. $V_e = \text{Speed of escalator}$

$V_c = \text{Speed of the child}$

$x = \text{Distance between 2 floors}$

We assume that it takes $t$ seconds for the child to reach the next floor.

So, $V_c = \frac{x}{t}$

We know that it takes 20 secs for the escalator to reach the next floor.

So, $V_e = \frac{x}{20}$.

We also know that it takes 4 seconds for the child on the escalator. So,

$V_e + V_c = \frac{x}{4}$

Now using these, we get

$V_e + V_c = \frac{x}{20}+ \frac{x}{t} = \frac{x}{4}$

You can solve the equation to find t now.